Product Measures and the Fubini Tonelli Theorems — Examples

Since $e^{-x^2}{e}^{-y^2}=e^{-(x^2+y^2)}$ is nonnegative, Tonelli gives $I^2={\int }_{\mathbb{R}}{\int }_{\mathbb{R}}{e}^{-(x^2+y^2)}dydx\mathrm{.}$

By [L2], the double integral in step 1.1 is $\pi$. Since $I\ge 0$, it follows that $I^2=\pi$ and hence $I=\sqrt{\pi }$.

For $(x,y)\in (0,1{)}^2$, [L1] gives $\frac{1}{1-xy}={\sum }_{n=0}^{\mathrm{\infty }}(xy{)}^n$ with nonnegative terms.

Tonelli and [L2] allow termwise integration: ${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy={\sum }_{n=0}^{\mathrm{\infty }}\left({\int }_0^1{x}^{n}dx\right)\left({\int }_0^1{y}^{n}dy\right)={\sum }_{n=0}^{\mathrm{\infty }}\frac{1}{(n+1{)}^2}\mathrm{.}$

Reindexing and applying [A1] yields ${\int }_0^1{\int }_0^1\frac{1}{1-xy}dxdy={\sum }_{m=1}^{\mathrm{\infty }}\frac{1}{m^2}=\frac{{\pi }^2}{6}\mathrm{.}$

Applying [L1] to $f(x)=x^2$ gives [L1] ${\lambda }_2(R)={\int }_0^1{x}^{2}dx\mathrm{.}$

Since $(x^3\mathrm{/}3{)}^{\mathrm{\prime }}=x^2$, [L2] gives [L2, step 1.1] ${\lambda }_2(R)={\int }_0^1{x}^{2}dx=\frac{1}{3}\mathrm{.}$ ∎

For $x\in [-1,1]$, the vertical section of $D$ is $D_x=(-\sqrt{1-x^2},\sqrt{1-x^2}),$ so ${\lambda }_1(D_x)=2\sqrt{1-x^2}$; outside $[-1,1]$ the section is empty. Applying [L1] to ${\mathbf{1}}_D$ therefore gives ${\lambda }_2(D)={\int }_{-1}^{1}2\sqrt{1-x^2}dx\mathrm{.}$

The same set $D$ is the unit ball in ${\mathbb{R}}^2$, so [L2] gives ${\int }_{-1}^{1}2\sqrt{1-x^2}dx={\lambda }_2(D)=\pi \mathrm{.}$

For each $x\in X$, the section $E_x=\{y:y<x\}$ is countable by the choice of $X$, hence measurable for $\mathcal{M}$. For each $y\in X$, the section $E^y=\{x:y<x\}$ has countable complement $\{x:x\le y\}$, hence is cocountable and measurable.

Suppose for contradiction that $E$ were product-measurable for $\mathcal{M}\otimes \mathcal{M}$. Since $\nu (X)=1$, the measure $\nu$ is finite and hence sigma-finite, so [L2] would give ${\int }_X\nu (E_x)d\nu ={\int }_X\nu (E^y)d\nu \mathrm{.}$ But step 1.1 makes $\nu (E_x)=0$ for every $x$ and $\nu (E^y)=1$ for every $y$, so the two sides are $0$ and $1$, a contradiction. Therefore $E$ is not product-measurable, even though all of its sections are measurable. Thus the displayed implication is false.

For fixed $x\in [0,1]$, the section $D_x=\{x\}$ has counting measure $1$, so ${\int }_X\nu (D_x)d\mu (x)={\int }_0^{1}1dx=1.$

For fixed $y\in [0,1]$, the section $D^y=\{y\}$ has Lebesgue measure $0$, so ${\int }_Y\mu (D^y)d\nu (y)={\int }_{[0,1]}0d\nu =0.$ The iterated integrals are unequal, so Tonelli fails once the counting-measure factor is not sigma-finite.

Direct differentiation gives $f(x,y)={\mathrm{\partial }}_y\left(\frac{y}{x^2+y^2}\right)=-{\mathrm{\partial }}_x\left(\frac{x}{x^2+y^2}\right)\mathrm{.}$

Integrating the first identity of step 1.1 in $y$ from $0$ to $1$ gives ${\int }_0^{1}f(x,y)dy=\frac{1}{1+x^2}\mathrm{.}$ Integrating in $x$ and applying [L1] yields ${\int }_0^1({\int }_0^{1}f(x,y)dy)dx={\int }_0^1\frac{dx}{1+x^2}=\frac{\pi }{4}\mathrm{.}$

Repeating the same calculation with the second identity of step 1.1 gives ${\int }_0^1({\int }_0^{1}f(x,y)dx)dy=-{\int }_0^1\frac{dy}{1+y^2}=-\frac{\pi }{4}\mathrm{.}$ Therefore the iterated integrals exist and are unequal, so the conclusion of Fubini fails. In particular $f\mathrm{\notin }{L}^1((0,1{)}^2)$, because otherwise Fubini's theorem for L^1 functions on a sigma-finite product would force them to agree.

The first copy of $g$ contributes the two iterated values of $f$, while the second copy contributes the same values with the order reversed. Therefore both iterated integrals of $g$ exist and are equal to $\pi \mathrm{/}4+(-\pi \mathrm{/}4)=0.$

The absolute integral of $g$ is the sum of the absolute integrals of the two copies, so it is still infinite because each copy carries the non-$L^1$ singularity of [L1]. Thus equal iterated integrals do not imply $L^1$-integrability.

By [L1], $\rho$ and $\tau$ agree on every measurable rectangle.

The same fact [L1, step 1.1] gives $\rho (D)=1\mathrm{\ne }0=\tau (D)$, so the two measures are distinct. Therefore the rectangle formula does not determine a unique product measure without sigma-finiteness.

AC implies [L8]: apply AC to the family $\{X_n:n\in \mathbb{N}\}$ of a sequence of nonempty sets and compose its choice function with $n\mapsto X_n$. This supplies countable choice for [L2], [L4] and [L5]. AC also supplies the selector in [L6]; fix that Vitali set $V\subseteq [0,1]$, which is non-Lebesgue-measurable, and set $E=\{0\}\times V$.

Both factors are sigma-finite by [L2]. The rectangle $Z=\{0\}\times [0,1]$ is product-measurable and [L3] and [L4] give $(\lambda \times \lambda )(Z)=\lambda (\{0\})\lambda ([0,1])=0\cdot 1=0$. Since $V\subseteq [0,1]$, we have $E\subseteq Z$.

If $E$ belonged to $\mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$, then [L1] would make $E_0=\{y:(0,y)\in E\}=V$ Lebesgue measurable, contrary to step 1.1. Thus the subset $E$ of the measurable null set $Z$ is not product-measurable. By [L7] the product is not complete, although [L5] makes both factors complete.

By [A1] and [L1], the indicator $f$ is measurable for the completed product measure.

The section at $0$ is $f_0={\mathbf{1}}_N$, whose support $N$ is not Lebesgue measurable. Hence $f_0$ is not measurable. So completed-product measurability gives section measurability only almost everywhere, not at every parameter.